When $|x| < \frac{1}{2}$,the coefficient of $x^4$ in the expansion of $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)}$ is

The partial fractions of $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4}$ are:

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$,then the value of $A+B+C+D+E+F=$

If the quotient and remainder obtained when the expression $3x^5-6x^4+2x^3+4x^2-5x+8$ is divided by the expression $x^2-2x+3$ are $ax^3+bx^2+cx+d$ and $px+q$ respectively,then $ab+cd=$

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

If $\frac{1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ and $\frac{x}{(x-1)(x-2)(x-3)}=\frac{P}{x-1}+\frac{Q}{x-2}+\frac{R}{x-3}$, then $A+2B+3C=$